There are two values of a for which the determinant, $Δ=\begin{vmatrix}1 & -2 & 5\\0 & a & -1\\0 & 4 & 2a\end{vmatrix}=86,$ then the sum of there values of a is :

$\sqrt{41}$ 0 $-\sqrt{41}$ $2\sqrt{41}$

0

The correct answer is Option (2) → 0 $Δ=\begin{vmatrix}1 & -2 & 5\\0 & a & -1\\0 & 4 & 2a\end{vmatrix}=86⇒2a^2+4=86$ so $a^2=41$ $a=±\sqrt{41}$ $∑a_i=0$